Problem 60 - Chemistry The Science in Context

Question

The hydrogen atomic emission spectrum includes a UV line with a wavelength of \(92.3 \mathrm{nm} .\) a. Is this line associated with a transition between different excited states or between an excited state and the ground state? b. What is the value of \(n_{1}\) of this transition? c. What is the energy of the longest wavelength photon that a ground-state hydrogen atom can absorb?

Step-by-Step Solution

Answer

a. The UV line with a wavelength of 92.3 nm is associated with a transition between different excited states, not between an excited state and the ground state. b. The value of \(n_{1}\) for this transition is 2. c. The energy of the longest wavelength photon that a ground-state hydrogen atom can absorb is \(1.635\times 10^{-18}\,\mathrm{J}\).

Step 1: Using the Rydberg Formula

The Rydberg formula describes the wavelengths \(\lambda\) of the hydrogen atomic emission spectrum as follows: $$\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$ Where \(R_H\) is the Rydberg constant for hydrogen, approximately equal to \(1.097\times 10^7 m^{-1}\), \(n_1\) is the initial energy level, and \(n_2>n_1\) is the final energy level after the transition. Given the wavelength \(\lambda = 92.3\,\mathrm{nm}=92.3\times 10^{-9}\,\mathrm{m}\), we can use the formula to find the difference between energy levels.

Step 2: Calculate the inverse of the wavelength difference

Using the given wavelength, calculate the left-hand side of the Rydberg equation: $$\frac{1}{\lambda}= \frac{1}{92.3\times 10^{-9}\,\mathrm{m}}=1.084\times 10^7\,\mathrm{m}^{-1}$$

Step 3: Compute the energy level difference

Now, use this value along with the Rydberg constant to find the difference of the energy levels: $$\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1.084\times 10^7\,\mathrm{m}^{-1}}{1.097\times 10^7\,\mathrm{m}^{-1}}=0.988$$

Step 4: Determine the type of transition

Since the difference between the energy levels \(n_1\) and \(n_2\) is less than 0.989, this indicates that the transition occurs between two different excited states and not between the ground state and an excited state.

Step 5: Calculate the initial energy level \(n_{1}\)

Now, we can find \(n_1\) by solving the equation: $$\frac{1}{n_{1}^2} = \frac{1}{n_2^2} - 0.988$$ It is known that \(n_2 > n_1\), so assuming \(n_1 = 1\), the smallest possible value for \(n_1\), we get \(n_2 = 2\). Then we have: $$\frac{1}{1^2} = \frac{1}{2^2} - 0.988$$ which leads to a negative result for \(n_2^2\). Therefore, \(n_1\) cannot be equal to 1. We find the correct value by trial and error. When \(n_1 = 2\), we get a valid result: $$\frac{1}{2^2} = \frac{1}{n_2^2} - 0.988 \Rightarrow n_2^2=16$$ So, \(n_1\) for this transition is \(2\).

Step 6: Calculate the energy of the longest wavelength photon

The longest wavelength photon that a ground-state hydrogen atom can absorb corresponds to the smallest energy gap, which occurs when the atom transitions from the ground state \(n_1=1\) to the first excited state \(n_2=2\). We can use the Rydberg formula with these values to find the wavelength and then use the energy-wavelength relationship to find the energy. The Rydberg formula becomes: $$\frac{1}{\lambda}=R_H \left(\frac{1}{1}-\frac{1}{2^2}\right)=\frac{3}{4}R_H$$ Now, find the wavelength: $$\lambda = \frac{1}{\frac{3}{4}R_H} = \frac{4}{3R_H}$$ Calculate the energy by using the relationship between energy and wavelength: $$E = \frac{hc}{\lambda}$$ Where \(h\) is Planck's constant, \(6.626\times 10^{-34}\,\mathrm{Js}\), and \(c\) is the speed of light, \(3\times 10^8\,\mathrm{m/s}\). Plugging in the values, we get: $$E = \frac{6.626\times 10^{-34}\,\mathrm{Js}\times 3\times 10^8\,\mathrm{m/s}}{\frac{4}{3\times 1.097\times 10^7\,\mathrm{m}^{-1}}} = 1.635\times 10^{-18}\,\mathrm{J}$$ The energy of the longest wavelength photon that a ground-state hydrogen atom can absorb is \(1.635\times 10^{-18}\,\mathrm{J}\).

Key Concepts

Rydberg Formula

To understand the hydrogen emission spectrum, we first need to talk about the Rydberg Formula, a crucial tool in analyzing atomic transitions. The Rydberg Formula is used to predict the wavelengths of light emitted by electrons in hydrogen as they move between energy levels. Here's the basic idea:

• The formula is given by \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( \lambda \) is the wavelength, \( R_H \) is the Rydberg constant for hydrogen, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final energy states, respectively.
• The Rydberg constant \( R_H \) is about \( 1.097 \times 10^7 \ \mathrm{m}^{-1} \).

In simple terms, electrons in a hydrogen atom emit or absorb energy when they jump from one energy level to another. The Rydberg Formula allows us to calculate the wavelength of this emitted or absorbed light. This formula is fundamental for understanding why hydrogen can emit different types of light, like UV or visible light. By plugging the wavelength of a known emission line into the formula, we can find the initial and final states of the electron responsible for that line.

Excited State Transitions

In the context of the hydrogen emission spectrum, excited state transitions refer to the movement of electrons between different energy levels, or orbits, within the atom. When electrons jump from a higher energy level to a lower one, they release energy in the form of light. Let’s break down what we mean by excited state transitions.

• An electron starts in a higher energy state and moves to a lower one, releasing a photon with energy corresponding to the difference in energy levels.
• If a transition occurs between two excited states, both states are above the lowest energy state (the ground state). For example, a move from \( n_2 = 4 \) to \( n_1 = 2 \).
• If the transition ends in the ground state \( n_1 = 1 \), it's a transition between an excited state and the ground state \( n_2 > 1 \).

By calculating energy level differences using the Rydberg Formula, we can identify whether a photon emission line, like the one at \( 92.3 \ \mathrm{nm} \), is an excited to excited or an excited to ground state transition. This identification is crucial because it tells us about the energy and dynamics of electron transitions within the atom.

Photon Energy Calculation

Calculating the energy of a photon is essential when talking about transitions because it tells us how much energy is involved in an electron's movement between different states. To find the energy of a photon, we generally use the formula:

• \( E = \frac{hc}{\lambda} \), where \( E \) is energy, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \ \mathrm{Js} \), \( c \) is the speed of light \( 3 \times 10^8 \ \mathrm{m/s} \), and \( \lambda \) is the wavelength of the photon.
• The energy calculated describes how much energy is released (in emission) or absorbed (in absorption) as the electron transitions between levels.

For example, if calculating the energy for the longest wavelength photon a ground-state hydrogen atom can absorb, we start with the Rydberg Formula to find the wavelength and then plug into the energy formula. This step helps reveal the specific energy characteristics of photons related to specific electronic transitions, providing a deeper insight into atomic behaviors and reactions.